Question

The value of \[ \frac{\sqrt{3}\,\cosec 20^\circ-\sec 20^\circ}{\cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ} \] is:

• A. 64
• B. 48
• C. 46
• D. 40

Hint:

Products like $\cos20^\circ\cos40^\circ\cos60^\circ\cos80^\circ$ can often be simplified using standard trigonometric product identities.

Answer 1

The Correct Option is A

Step 1: Simplify the numerator.
\[ \sqrt{3}\,\cosec 20^\circ-\sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ}-\frac{1}{\cos 20^\circ} \] \[ = \frac{\sqrt{3}\cos 20^\circ-\sin 20^\circ}{\sin 20^\circ \cos 20^\circ} \] Step 2: Use the identity.
\[ \sqrt{3}\cos 20^\circ-\sin 20^\circ = 2\cos(20^\circ+30^\circ)=2\cos 50^\circ \] Hence, \[ \text{Numerator}=\frac{2\cos 50^\circ}{\sin 20^\circ \cos 20^\circ} \] Step 3: Simplify the denominator.
\[ \cos 20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ = \frac{1}{4}\sin 20^\circ \] Step 4: Compute the value.
\[ \frac{\frac{2\cos 50^\circ}{\sin 20^\circ \cos 20^\circ}} {\frac{1}{4}\sin 20^\circ} = 8\cdot\frac{\cos 50^\circ}{\cos 20^\circ} \] Since $\cos 50^\circ=\sin 40^\circ=2\sin 20^\circ\cos 20^\circ$, \[ = 8\times 8=64 \]